What is the center of a circle whose equation is $x^2+y^2-12 x-2 y+12=0 ?$
A. $(-12,-2)$
B. $(-6,-1)$
C. $(6,1)$
D. $(12,2)$
Answer (1)
To find the center of the circle given by the equation x 2 + y 2 − 12 x − 2 y + 12 = 0 , we complete the square for both x and y terms. This allows us to rewrite the equation in the standard form ( x − h ) 2 + ( y − k ) 2 = r 2 , where ( h , k ) is the center of the circle. The steps are as follows:
Complete the square for x : x 2 − 12 x becomes ( x − 6 ) 2 − 36 .
Complete the square for y : y 2 − 2 y becomes ( y − 1 ) 2 − 1 .
Substitute these back into the original equation and simplify: ( x − 6 ) 2 − 36 + ( y − 1 ) 2 − 1 + 12 = 0 , which simplifies to ( x − 6 ) 2 + ( y − 1 ) 2 = 25 .
Identify the center of the circle from the standard form: ( h , k ) = ( 6 , 1 ) .
Therefore, the center of the circle is ( 6 , 1 ) .
Explanation
Analyze the problem We are given the equation of a circle: x 2 + y 2 − 12 x − 2 y + 12 = 0 . Our goal is to find the center of this circle. To do this, we will rewrite the equation in the standard form of a circle, which is ( x − h ) 2 + ( y − k ) 2 = r 2 , where ( h , k ) is the center and r is the radius.
Complete the square for x terms First, we complete the square for the x terms. We have x 2 − 12 x . To complete the square, we take half of the coefficient of the x term, which is − 12/2 = − 6 , and square it: ( − 6 ) 2 = 36 . So, we can rewrite x 2 − 12 x as ( x − 6 ) 2 − 36 .
Complete the square for y terms Next, we complete the square for the y terms. We have y 2 − 2 y . To complete the square, we take half of the coefficient of the y term, which is − 2/2 = − 1 , and square it: ( − 1 ) 2 = 1 . So, we can rewrite y 2 − 2 y as ( y − 1 ) 2 − 1 .
Substitute back into original equation Now, we substitute these back into the original equation: ( x − 6 ) 2 − 36 + ( y − 1 ) 2 − 1 + 12 = 0 .
Simplify the equation We simplify the equation: ( x − 6 ) 2 + ( y − 1 ) 2 = 36 + 1 − 12 = 25 .
Identify the center Now the equation is in the standard form ( x − 6 ) 2 + ( y − 1 ) 2 = 25 . From this, we can identify the center of the circle as ( h , k ) = ( 6 , 1 ) .
State the final answer Therefore, the center of the circle is ( 6 , 1 ) .
Examples
Understanding the equation of a circle is useful in many real-world applications. For example, when designing a circular garden, knowing the center and radius helps in planning the layout and determining the amount of fencing needed. Similarly, in architecture, circular arches and domes require precise calculations based on the circle's equation to ensure structural integrity and aesthetic appeal. In navigation, the concept of circles is used to determine the range of a radar or the area covered by a radio signal, where the transmitter is at the center of the circle.
